Answer:
A. To find the derivative of h(x), we can use the chain rule:
h(x) = a(-2x + 1)^5 - b
h'(x) = a * 5(-2x + 1)^4 * (-2) = -10a(-2x + 1)^4
To find the second derivative, we can again use the chain rule:
h''(x) = -10a * 4(-2x + 1)^3 * (-2) = 80a(-2x + 1)^3
B. To show that h is monotonic, we need to show that h'(x) is either always positive or always negative. Since h'(x) is a multiple of (-2x + 1)^4, which is always non-negative, h'(x) is always either positive or negative depending on the sign of a. If a > 0, then h'(x) is always negative, which means that h(x) is decreasing. If a < 0, then h'(x) is always positive, which means that h(x) is increasing.
C. To find the critical points, we need to find where h'(x) = 0:
h'(x) = -10a(-2x + 1)^4 = 0
-2x + 1 = 0
x = 1/2
Thus, the critical point is at x = 1/2. This value is independent of a and b, as neither a nor b appear in the calculation of the critical point.